Question #152230

A charge moves a distance of 15.0 cm in the direction of a uniform electric field whose magnitude is 195 N/C. As the charge moves, its electrical potential energy increases by 7.5 × 10−6 J. What is the charge on the moving particle?


1
Expert's answer
2020-12-21T11:00:47-0500

By the definition, electrical potential energy is the energy that is needed to move a charge against an electric field. The change in electrical potential energy of the charge can be found as follows:


ΔU=UfUi=W,\Delta U=U_f-U_i=-W,ΔU=W=qEd,\Delta U=-W=-qEd,q=ΔUEd=7.5106 J195 NC0.15 m=2.56107 C.q=-\dfrac{-\Delta U}{Ed}=-\dfrac{-7.5\cdot10^{-6}\ J}{195\ \dfrac{N}{C}\cdot 0.15\ m}=2.56\cdot10^{-7}\ C.

Answer:

q=2.56107 C.q=2.56\cdot10^{-7}\ C.


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